Chapter 9.2, Problem 4CE

To determine

To prove : Tangent PA is congruent to tangent PB.

Explanation of Solution

Given information:

  

If the line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangents.

Then,

  $\begin{array}{l}OA\perp PA\hfill \\ OB\perp PB\hfill \end{array}$

Then once can say that,

  $\begin{array}{l}\angle PAO=90°\hfill \\ \angle PBO=90°\hfill \end{array}$ It is required to prove two triangles are congruent.

Consider in ?PAO and ?PBO,

  $OA=OB$ (Radius of a circle is equal)

  $\angle PAO=\angle PBO$ (Both the angles are 90°)

  $PO=OP$ (Common side of a triangle)

So, it implies that

  $\Delta PAO\cong \Delta PBO$ (Side-Angle-Side Rule)

Since the corresponding parts of congruent triangles are congruent.

Therefore, the following holds good PA and PB.